Finding Volume of a Bounded Cylinder with Double Integral

[One-D]

problem:
find volume bordered by cylinder x^2 + y^2 = 4 and y+z=4 and z=4.
the answer is said to be 16p. but I couldn't find it.
I found it in double integral part.so it must be solved with double integral. I tried with Jacobian tranformation. nut still couldn't solve it. I was confused with what function should be integrated. because the cylinder has no border in z line.

 

[HallsofIvy]

Start, of course, by drawing a picture. I drew the picture using y as horizontal axis, z a vertical axis. The cylinder is just the 2 lines y= 2, y= -2. y+ z= 4 is represented by a line from (-2, 6) to (0,4) to (2,2). Notice that the plane z= 4 cuts that line in half. I interpret the "volume bordered by ..." as meaning the sections both below and above z= 4. Since they are clearly symmetric, just find the area of the section below z= 4 and double.

Projecting the figure down into the xy-plane, we get the portion of the circle x²+ y²= 4 above the x-axis. x ranges from -2 to 2 and, for each x, y ranges from 0 up to √(4- x²). For each (x,y) in that half circle, z ranges from z= 4-y (below) to 4 (above). The volume is given by
\int_{-2}^2\int_0^{\sqrt{4-x^2}}y dy dx

If you've done integrals in polar coordinates, since this is a half circle, you could also do it as
\int_0^2 \int_0^{\pi} r sin(\theta) r d\theta dr.

I don't know what "16p" is. What is "p"? I get 16/3 as the volume.

 

[One-D]

yesterday I had thought it again and I think I have found it. but it's seems diffrent from yours. why does the integration to y? shouldn't it (4-y)? I assumed that the section is under z=4 and above z+y=4 inside x^2+y^2=4.
p=3.14. I will check it again. thanx
oh yes I got it now. you subtract z=4 with z=4-y. am I right? thanx

 

[HallsofIvy]

Yes, that is correct.